Newton's Method tahn(x)

1. Feb 4, 2013

math199711

I was hoping someone could help me out with this problem. I need to find the values, that Newton's method converges for tanh(x).

So far I set up the algorithm:

x$\tiny_{k+1}$ = x$_{k}$ - $\frac{tanhx}{sech^{2}x}$

I simplified it to:

x$\tiny_{k+1}$ = x$_{k}$ - ($\frac{1}{2}$sinh(2x))

And then I took the derivative to see when the absolute value of the derivative was less than one.

|1 - cosh(2x)| < 1

I did all of this and came up with the interval of convergence as being:

$\frac{1}{2}$ln(2$\pm \sqrt{3}$)

These numbers are approximately .-658 and .658 respectively.

However when I test out 1, it converges using Newton's method, yet it falls out of the range that I determined.

Any help on the problem would be appreciated.